A secant line is a line drawn through two points on a curve. The Mean Value Theorem relates the slope of a secant line to the slope of a tangent line.

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1 The Mean Value Theorem A secant line is a line drawn through two points on a curve. The Mean Value Theorem relates the slope of a secant line to the slope of a tangent line. The Mean Value Theorem. If f is continuous on a x b and differentiable on a < x < b, there is a number c in a < x < b such that f(b) f(a) = f (c). b a The picture below shows why this makes sense. I ve drawn a secant line through the points (a, f(a)) and (b, f(b)). The Mean Value Theorem says that somewhere in between a and b, there is a point c on the curve where the tangent line has the same slope as the secant line. tangent line (b,f(b)) secant line (a,f(a)) a c b Lines with the same slope are parallel. To find a point where the tangent line is parallel to the secant line, take the secant line and slide it (without changing its slope) until it s tangent to the curve. If you experiment with some curves, you ll find that it s always possible to do this (provided that the curve is continuous and differentiable as stipulated in the theorem). Example. Consider f(x) = x 3 + 3x on the interval 5 x 1. Since f is a polynomial, f is continuous on 5 x 1 and differentiable on 5 < x < 1. f(1) f( 5) 1 ( 5) = 4 ( 50) 1 ( 5) = 9, so I should be able to find a number c between 5 and 1 such that f (c) = 9. f (x) = 3x + 6x, so f (c) = 3c + 6c. Set f (c) equal to 9 and solve for c: 3c + 6c = 9, c + c = 3, c + c 3 = 0, (c + 3)(c 1) = 0, c = 3 or c = 1. c = 11 is not in the interval 5 < x < 1 it s an endpoint but c = 3 is. c = 3 is a number satisfying the conclusion of the Mean Value Theorem. Notes: 1. There may be more than one value for c which works.. In general, finding a value of c that works may be difficult. The theorem only guarantees that such a c exists. 1

2 You have to ensure that the hypotheses of the theorem are satisfied before you apply it. Example. Consider f(x) = 1 on the interval 1 x 1. Then x f(1) f( 1) 1 ( 1) However, f (x) = x 3, and f (c) = = 0 has no solution. c3 This does not contradict the Mean Value Theorem, because f blows up at x = 0, which is in the middle of the interval 1 x 1. = 0. Example. Calvin Butterball runs a 100 yard dash in 0 seconds. Assume that the function s(t) which gives his position relative to the starting line is continuous and differentiable. Show that Calvin must have been running at 5 yards per second at some point during his run. When t = 0, he s at the starting line, so s = 0. When t = 0, he s at the finish line, so s = 100. Applying the Mean Value Theorem to s for 0 t 0, I find that there is a point c between 0 and 0 such that s (c) = = 5. That is, Calvin s velocity at t = c was 5 yards per second, which is what I wanted to show. The Mean Value Theorem is often used to prove mathematical results. Here s an example. You know that the derivative of a constant is zero. The converse is also true. Theorem. If f is continuous on the closed interval [a, b] and f (x) = 0 for all x in the open interval (a, b), then f is constant on the closed interval [a, b]. Proof. To prove this, let d be any number such that a < d b. The Mean Value Theorem applies to f on the interval [a, d], so there is a number c such that a < c < d and By assumption, f (c) = 0. Therefore, f(d) f(a) d a = f (c). f(d) f(a) d a = 0, so f(d) f(a) = 0, and f(d) = f(a). Since d was an aribtrary number such that a < d b, it follows that f(a) = f(x) for all x in [a, b]. This means that f is constant on the interval. Example. I know that d dx x3 = 3x. If f(x) is any other function such that d dx f(x) = 3x, then d ( ) f(x) x 3 = 3x 3x = 0. dx By the theorem, f(x) x 3 = c, where c is a constant. Therefore, f(x) = x 3 + c. In other words, the only functions whose derivatives are 3x are functions like x 3, x , x 3 7, and so on.

3 When I discuss antiderivatives later on, I ll express this fact by writing 3x dx = x 3 + c. In the case where the Mean Value Theorem applies and f(a) = f(b), I get f(b) f(a) b a = 0. The MVT says there is a point c in a < x < b such that f (c) = 0. This is called Rolle s Theorem, and a special case may be stated more informally as follows: For a nice function, there is at least one horizontal tangent between every pair of roots. a b In the picture above, there are three critical points between the roots at a and b. Example. By the Mean Value Theorem, the function f(x) = x(x 0)(x 00)(x 000) has critical points places where f = 0 between 0 and 0, between 0 and 00, and between 00 and 000. Example. Prove that the function f(x) = x 5 + 7x x 18 has exactly one root. Note that a graph is not a proof! Since f(10) = and f( 10) = , and since f is continuous, the Intermediate Value Theorem implies that there is a root between 10 and 10. Thus, f has at least one root. Suppose that f has more than one root. Suppose, in particular, that a and b are distinct roots of f. By Rolle s theorem, f must have a horizontal tangent between a and b. However, the derivative is f (x) = 5x 4 + 1x Since the powers of x are even, f (x) > 0 for all x: There are no horizontal tangents. This contradiction shows that there can t be more than one root. Since I already know that there s at least one root, there must be exactly one root. Example. Here is another mathematical result which follows from the Mean Value Theorem; it will be useful in graphing curves. If f is differentiable on a < x < b and f (x) > 0 on a < x < b, then f increases on a < x < b. To say that f increases means that f goes up from left to right. 3

4 To see this, take p and q between a and b; say a < p < q < b. I want to show f(p) < f(q). By the MVT, f(q) f(p) q p = f (c) for some c between p and q. But f (c) > 0, so f(q) f(p) q p This proves that f increases on the interval. > 0, f(q) f(p) > 0, f(q) > f(p). Example. Here s another example of how the Mean Value Theorem can be used to prove a mathematical result in this case, an inequality. Apply the MVT to f(x) = tan x on the interval 0 x k, where k < π to avoid running into the vertical asymptote. I get for some c between 0 and k. Now (sec c) 1 and tan 0 = 0, so tan k tan 0 k 0 = (sec c) tan k k 1 and hence tan k k. A picture which illustrates this (not to scale) follows: The curve is the graph of y = tan x and the line is y = x. You can see that the curve appears to lie above the line. Example. (Using the Mean Value Theorem to estimate a function value) Suppose that f is a differentiable function, f(3) =, and 3 f (x) 4 for all x. Prove that 8 f(5) 10. Apply the Mean Value Theorem to f on the interval 3 x 5: f(5) f(3) 5 3 = f (c) where 3 < c < 5. c 005 by Bruce Ikenaga 4

5 Then since 3 f (c) 4, I have f(5) f(3) f(5) f(5) f(5) 10 c 005 by Bruce Ikenaga 5